Problem: Baby Ehab was toying around with arrays. He has an array a of length n. He defines an array to be good if there's no way to partition it into 2 subsequences such that the sum of the elements in the first is equal to the sum of the elements in the second. Now he wants to remove the minimum number of elements in a so that it becomes a good array. Can you help him?

A sequence b is a subsequence of an array a if b can be obtained from a by deleting some (possibly zero or all) elements. A partitioning of an array is a way to divide it into 2 subsequences such that every element belongs to exactly one subsequence, so you must use all the elements, and you can't share any elements.

Input

The first line contains an integer n (2 ≤ n ≤ 100) — the length of the array a.

The second line contains n integers a_1, a_2, …, a_{n} (1 ≤ a_i ≤ 2000) — the elements of the array a.

Output

The first line should contain the minimum number of elements you need to remove.

The second line should contain the indices of the elements you're removing, separated by spaces.

We can show that an answer always exists. If there are multiple solutions, you can print any.

Examples

Input


4
6 3 9 12


Output


1
2

Input


2
1 2


Output


0

Note

In the first example, you can partition the array into [6,9] and [3,12], so you must remove at least 1 element. Removing 3 is sufficient.

In the second example, the array is already good, so you don't need to remove any elements.

Solution (in CPP):

Solution:
/*
                                               THE GRIMMEST OF Aint REAPER
                                                DEATH SHAint BE UPON YOU
                                                             .""--.._
                                                           []      `'--.._
                                                           ||__           `'-,
                                                         `)||_ ```'--..       \
                                     _                    /|//}        ``--._  |
                                  .'` `'.                /////}              `\/
                                 /  .""".\              //{///
                                /  /_  _`\\            // `||
                                | |(_)(_)||          _//   ||
                                | |  /\  )|        _///\   ||
                                | |L====J |       / |/ |   ||
                               /  /'-..-' /    .'`  \  |   ||
                              /   |  :: | |_.-`      |  \  ||
                             /|   `\-::.| |          \   | ||
                           /` `|   /    | |          |   / ||
                         |`    \   |    / /          \  |  ||
                        |       `\_|    |/      ,.__. \ |  ||
                        /                     /`    `\ ||  ||
                       |           .         /        \||  ||
                       |                     |         |/  ||
                       /         /           |         (   ||
                      /          .           /          )  ||
                     |            \          |             ||
                    /             |          /             ||
                   |\            /          |              ||
                   \ `-._       |           /              ||
                    \ ,//`\    /`           |              ||
                     ///\  \  |             \              ||
                    |||| ) |__/             |              ||
                    |||| `.(                |              ||
                    `\\` /`                 /              ||
                       /`                   /              ||
                      /                     |              ||
                     |                      \              ||
                    /                        |             ||
                  /`                          \            ||
                /`                            |            ||
                `-.___,-.      .-.        ___,'            ||
                         `---'`   `'----'`
*/
#include <bits/stdc++.h>
//#include<cstdio>
//#include<cstring>
//#include<algorithm>
//#include<iostream>
//#include<string>
//#include<vector>
//#include<stack>
//#include<bitset>
//#include<cstdlib>
//#include<cmath>
//#include<set>
//#include<list>
//#include<deque>
//#include<map>
//#include<queue>

#define rep(i, a, n) for (int  i=a;i<=n;i++)
#define per(i, a, n) for (int  i=a;i>=n;i--)
#define IOS ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
//#pragma GCC optimize(3)//O3
//#pragma GCC optimize(2)//O2
//#define endl "\n"
#define Mod(x) x % mod
#define toNum(x) x -'0'
#define fi first
#define se second
#define mm(a, b) memset(a,b, sizeof(a));
#define dbg(x) cout << #x << "=" << x << endl;
using namespace std;
typedef long long ll;
//typedef unsigned long long ull;
typedef pair<int, int> PII;
const int maxn = 3e6 + 10;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
#define cf int  t;cin>>t;while (t--) solve();
#define tty template <typename T>
tty T gcd (T a, T b) {while (b ^= a ^= b ^= a %= b);  return a;}
inline char nc() {static char buf[1000000], *p1 = buf, *p2 = buf;return p1 == p2 && (p2 = (p1 = buf) + fread (buf, 1, 1000000, stdin), p1 == p2) ? EOF : *p1++;}
//#define nc getchar
tty inline void read(T &sum) {char ch = nc();T tf = 0;sum = 0;while((ch < '0' || ch > '9') && (ch != '-')) ch = nc();tf = ((ch == '-') && (ch = nc()));while(ch >= '0' && ch <= '9') sum = sum * 10+ (ch - 48), ch = nc();(tf) && (sum = -sum);}
tty inline void write(T x){if(x==0) {putchar('0');return;} char F[200];T tmp=x>0?x:-x ;if(x<0)putchar('-') ;T cnt=0 ;while(tmp>0){F[cnt++]=tmp%10+'0';tmp/=10;}while(cnt>0)putchar(F[--cnt]) ;}

//大随机数
tty inline T ra(T z,T y){mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());return uniform_int_distribution<T>(z, y)(rng);}

//exgcd
tty T niyuan(T a, T b, T &x, T &y) {if (!b) {x = 1,y=0;return a;}T d = niyuan(b, a % b, x, y);T t = x;x=y;y=t-(a/b)*y;return d;}
tty T ax_bmodn(T a,T b,T n,T &g,T &x,T &y) {if (!n) { g = a, x = 1, y = 0; }else {ax___bmodn(n, a % n, g, y, x);y -= x * (a / n);}return x *= (b / g);}

//mod
tty inline T dmod(const string& s,T mod){T len = s.size();T ans = 0;for (T i = 0; i < len; i++)ans = (int)(((long long)ans * 10 + s[i] - '0') % mod); return ans;}
tty inline T ksc(T a,T b,T mod){T ans = 0;while( b > 0 ){if( b&1 ) ans = (ans + a) % mod;a = ( a + a ) % mod;b >>= 1;}return ans;}
tty inline T ksm(T x,T n,T mod){T ans = 1;while (n > 0){if (n & 1)ans = ans*x%mod;x = x*x%mod;n >>= 1;}return ans;}
tty inline T mod_jia(T a,T b,T mod){return (a%mod+b%mod)%mod;}
tty inline T mod_jian(T a,T b,T mod){return (a%mod+mod-b%mod)%mod;}
tty inline T mod_cheng(T a,T b,T mod){return (a%mod*b%mod)%mod;}
tty inline T C(T n,T m,T mod) {if (m > n) return 0;T ans = 1;for (T i = 1; i <= m; i++) {T a = (n + i - m) % mod;T b = i % mod;ans = ans * (a * ksm(b, mod - 2,mod) % mod) % mod;}return ans;}

//RMQ
tty inline void rmqmin(T f[][20],T n){for (int i = 1; i <= n; i++)cin >> f[i][0];for (int j = 1; j <= 20; j++)for (int i = 1; i <= n; i++)if (i + (1 << j) - 1 <= n)f[i][j] = min(f[i][j - 1], f[i + (1 << (j-1))][j - 1]);}
tty inline T zymin(T f[][20],T z,T y){int x=int (log(y-z+1)/log(2));return min(f[z][x],f[y-(1<<x)+1][x]);}
tty inline void rmqmax(T f[][20], T n) {for (int i = 1; i <= n; i++)cin >> f[i][0];for (int j = 1; j <= 20; j++)for (int i = 1; i <= n; i++)if (i + (1 << j) - 1 <= n)f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);}
tty inline T zymax(T f[][20],T z,T y){int x=int (log(y-z+1)/log(2));return max(f[z][x],f[y-(1<<x)+1][x]);}

//旋转公式
tty inline T rotatex(T x,T y,T ang,T ox=0,T oy=0){return (x-ox)*cos(ang)-(y-oy)*sin(ang)+ox;}
tty inline T rotatey(T x,T y,T ang,T ox=0,T oy=0){return (x-ox)*sin(ang)+(y-oy)*cos(ang)+oy;}

//快速幂
ll qpow(ll x,ll y) {ll ans=1;while(y){if(y&1)ans=ans*x%mod;y=y/2;x=x*x%mod;}return ans;}
//-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

int n, m, sum, cnk,vis, maxx, minn = INF, u, l, r, pre, op, p, k, z, ok, x, y, temp, q, sum1, sum2, A, B, cnt, ask, t, nums, len, b;
int a[2010];
int dp[101][200001];
void solve()
{
    cin >> n;
    rep(i,1,n){
        cin >> a[i];
        sum+=a[i];
        for(int j=a[i];j<=2000*n;j++) {
            dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - a[i]] + a[i]);
        }
    }
    if((sum&1) || (dp[n][sum/2]!=sum/2)) {
        cout<<0;
        return;
    }
    cout<<1<<endl;
    rep(i,1,n){
        if(a[i]&1 || dp[n][(sum-a[i])/2]!=(sum-a[i])/2){
            cout<<i;
            return;
        }
    }
    return;
}
signed main()
{

//    cf;
    solve();
    return 0;
}




//
//friend bool operator <(node s1,node s2){
//    return s1.x>s2.x;
//}